smc-1211-70-Projections

Given a non-zero vector \(\left(\begin{array}{l}p \\ q\end{array}\right)\), it is known that the vector \(\left(\begin{array}{c}q \\ -p\end{array}\right)\) is perpendicular to \(\left(\begin{array}{l}p \\ q\end{array}\right)\) and has the same magnitude. (Do NOT prove this.)
Points \(A\) and \(B\) have position vectors \(\overrightarrow{O A}=\left(\begin{array}{l}a_1 \\ a_2\end{array}\right)\) and \(\overrightarrow{O B}=\left(\begin{array}{l}b_1 \\ b_2\end{array}\right)\), respectively.
Using the given information, or otherwise, show that the area of triangle \(O A B\) is \(\dfrac{1}{2}\left|a_1 b_2-a_2 b_1\right|\). (3 marks)

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The point \(P\) lies on the circle centred at \(I(r, 0)\) with radius \(r>0\), such that \(\overrightarrow{I P}\) makes an angle of \(t\) to the horizontal.
The point \(Q\) lies on the circle centred at \(J(-R, 0)\) with radius \(R>0\), such that \(\overrightarrow{J Q}\) makes an angle of \(2 t\) to the horizontal.

Note that \(\overrightarrow{O P}=\overrightarrow{O I}+\overrightarrow{I P}\) and \(\overrightarrow{O Q}=\overrightarrow{O J}+\overrightarrow{J Q}\).
Using part (i), or otherwise, find the values of \(t\), where \(-\pi \leq t \leq \pi\), that maximise the area of triangle \(O P Q\). (4 marks)

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♦♦♦ Mean mark (i) 18%.

\(\Big{\|}\text{proj}_{\overrightarrow{OA^{′}}} \overrightarrow{OB} \Big{\|}\)	\(=\ \text{⊥ height of}\ \triangle OAB\ \text{from side}\ \overrightarrow{OA} \)
	\(= \Bigg{\|} \dfrac{\overrightarrow{OB} \cdot \overrightarrow{OA^{′}}}{\|\overrightarrow{OA^{′}}\|} \Bigg{\|} \)
	\(= \Bigg{\|} \dfrac{b_1a_2-b_2a_1}{\|\overrightarrow{OA^{′}}\|} \Bigg{\|} \)

\(\text{Area}\ \triangle AOB\)	\(=\dfrac{1}{2} \Bigg{\|} \dfrac{b_1a_2-b_2a_1}{\|\overrightarrow{OA^{′}}\|} \Bigg{\|} \cdot \|\overrightarrow{OA} \| \)
	\(=\dfrac{1}{2}\left\|a_1 b_2-a_2 b_1\right\|\ \ \ (\text{noting}\ \ \|\overrightarrow{OA} \|=\|\overrightarrow{OA^{′}} \|) \)

ii.	\(\overrightarrow{OP}\)	\(=\overrightarrow{OI}+\overrightarrow{IP}\)
		\(= \left(\begin{array}{l}r \\ 0\end{array}\right) + \left(\begin{array}{c} r\ \cos \ t \\ r\ \sin \ t\end{array}\right) \)
		\(= \left(\begin{array}{c} r(1+ \cos \ t) \\ r\ \sin \ t\end{array}\right) \)

♦♦♦ Mean mark (ii) 8%.

\(\overrightarrow{OQ}\)	\(=\overrightarrow{OJ}+\overrightarrow{JQ}\)
	\(= \left(\begin{array}{c} -R \\ 0 \end{array}\right) + \left(\begin{array}{l} R\ \cos\ 2t \\ R\ \sin\ 2t\end{array}\right) \)
	\(= \left(\begin{array}{c} R(\cos\ 2t-1) \\ R\ \sin\ 2t\end{array}\right) \)

\(\text{Using part (i):}\)

\(A_{\triangle OPQ}\)	\(=\dfrac{1}{2} \big{\|} r(1+\cos\ t) \cdot R \sin\ 2t-r\ \sin\ t\ \cdot R(\cos\ 2t-1)\ \big{\|} \)
	\(=\dfrac{rR}{2} \big{\|} \sin\ 2t+\cos\ t\ \sin\ 2t-\sin\ t\ \cos\ 2t+\sin\ t\ \big{\|} \)
	\(=\dfrac{rR}{2} \big{\|} \sin\ 2t+\sin\ t\ + \sin(2t-t) \big{\|} \)
	\(=\dfrac{rR}{2} \big{\|} 2\sin\ t\ \cos\ t +2\sin\ t \big{\|} \)
	\(= rR \big{\|} \sin\ t(\cos\ t+1) \big{\|} \)

\(\text{Let}\ \ f(t)=\sin\ t(\cos\ t+1) \)

\(\text{SP’s when}\ \ f^{′}(t)=0:\)

\(\cos\ t\)	\(=\dfrac{1}{2} \)	\(\cos\ t\)	\(=-1\)
\(t\)	\(=\dfrac{\pi}{3}, \ – \dfrac{\pi}{3} \)	\(t\)	\(=\pi, \ -\pi \)

\(\text{Testing SP’s:}\)

\(f(\pi) = f(- \pi) = 0\)

\(\therefore A_{\triangle AOB}\ \text{is maximum when}\ \ t=\dfrac{\pi}{3}, \ – \dfrac{\pi}{3} \)

\(\Big{\|}\text{proj}_{\overrightarrow{OA^{′}}} \overrightarrow{OB} \Big{\|}\)	\(=\ \text{⊥ height of}\ \triangle OAB\ \text{from side}\ \overrightarrow{OA} \)
	\(= \Bigg{\|} \dfrac{\overrightarrow{OB} \cdot \overrightarrow{OA^{′}}}{\|\overrightarrow{OA^{′}}\|} \Bigg{\|} \)
	\(= \Bigg{\|} \dfrac{b_1a_2-b_2a_1}{\|\overrightarrow{OA^{′}}\|} \Bigg{\|} \)

\(A_{\triangle OPQ}\)	\(=\dfrac{1}{2} \big{\|} r(1+\cos\ t) \cdot R \sin\ 2t-r\ \sin\ t\ \cdot R(\cos\ 2t-1)\ \big{\|} \)
	\(=\dfrac{rR}{2} \big{\|} \sin\ 2t+\cos\ t\ \sin\ 2t-\sin\ t\ \cos\ 2t+\sin\ t\ \big{\|} \)
	\(=\dfrac{rR}{2} \big{\|} \sin\ 2t+\sin\ t\ + \sin(2t-t) \big{\|} \)
	\(=\dfrac{rR}{2} \big{\|} 2\sin\ t\ \cos\ t +2\sin\ t \big{\|} \)
	\(= rR \big{\|} \sin\ t(\cos\ t+1) \big{\|} \)

Vectors, EXT1 V1 2023 HSC 14c